## anonymous 4 years ago find the domain of the function f(x)= 1/sqrt (4x-8) please show calculations

1. anonymous

huh

2. anonymous

callum can you show the steps please

3. anonymous

(2, infinity)

4. anonymous

$f(x) = \frac{1}{\sqrt{4x-8}}$ now you want to look at conditions on 4x-8. What can x reasonably be? Well you cannot take a square root of a negative number, and the denominator cannot be zero, so we need: $4x-8 > 0$ $x>\frac{1}{2}$ So I was wrong above, the domain is actually: $\left(\frac{1}{2}, \infty \right)$

5. anonymous

x > 2

6. anonymous

think you divided the wrong side there

7. anonymous

oh yes sorry it's 2 lol, quite right. $(2,\infty)$

8. anonymous

9. anonymous

you'll have imaginary numbers if the square root is allowed to go negative

10. anonymous

thus not part of the real number system

11. anonymous

callum was right in his calculation 4x - 8 > 0 4(x-2) > 0 x-2 > 0 x > 2

12. anonymous

yeah my reasoning was right, it's just I divided the inequality wrong and got x>1/2. rickjbr corrected this to x>2

13. anonymous

thanks for the help to both of your

14. anonymous

np

15. anonymous

16. anonymous

The question reads Find and simplify the difference quotient f(x+h)-f(x)/h for the given function f(x)=2x^(2)-3x+5

17. anonymous

ah, right

18. anonymous

(2(x+h)^2-3(x+h)+5-2x^2+3x-5)/h

19. anonymous

(2(x^2+2xh+h^2)-3(x+h)+5-2x^2+3x-5)/h

20. anonymous

(2x^2+4xh+2h^2-3x-3h+5-2x^2+3x-5)/h

21. anonymous

(2h^2+4xh+3x)/h

22. anonymous

23. anonymous

checking, a bit rusty

24. anonymous

my options are a)4x+2h, b)4x+2h-3, c)4h+2x-3, d)4h+2x, e)4x+4h

25. anonymous

i'll redo it, might've missed something

26. anonymous

ok

27. anonymous

ah, silly

28. anonymous

that last step was it. instead of (2h^2+4xh+3x)/h it should be (2h^2+4xh-3h)/h

29. anonymous

simply factor out an h

30. anonymous

[h(2h+4x-3)]/h =2h+4x-3

31. anonymous

that's it

32. anonymous

thanks greatly. i dont know if you have the time...but im reviewing for a test and I have a whole lot of other questions. if you can't help me, i understand, i will just post on the main forum

33. anonymous

probably better off posting on main forum, if i can help, i'll try

34. anonymous

ok thanks for the help

35. anonymous

Find $\frac{f(x+h)-f(x)}{h}$ for $f(x)=2x^2-3x+5$ $\frac{f(x+h)-f(x)}{h} = \frac{(2(x+h)^2-3(x+h)+5) - (2x^2-3x+5)}{h}$ $= \frac{(2x^2+4xh+2h^2-3x-3h)-(2x^2-3x)}{h}$ $=\frac{4xh+2h^2-3h}{h} = 4x+2h-3$ We can verify that this is right by noting that $f'(x) = \lim_{h\rightarrow 0}\frac{f(x+h) - f(x)}{h}$ The left hand side of this is equal to the derivative of f, which we can just do right here: $f'(x) = 4x - 3$ and we should get that this equals the limit as h goes to zero of $\frac{f(x+h) - f(x)}{h}.$ We have already calculated that $\frac{f(x+h) - f(x)}{h} = 4x+2h-3$ so taking the limit of this as h goes to zero we get: $\lim_{h\rightarrow 0}(4x+2h - 3) = 4x+0-3 = 4x-3.$ Hence we ave verified that the simplification has been done correctly.